[proofplan] We compute the entries of the point Gram matrix $NN^\top$ directly from the incidence axioms of the [$2$-$(v,k,\lambda)$ design](/page/Block%20Design) and its [incidence matrix](/page/Incidence%20Matrix). This gives the matrix identity $NN^\top = (r-\lambda)I_v+\lambda J$, where $J$ is the all-one matrix. Let $\mathbb{1}:=(1,\dots,1)^\top\in\mathbb{R}^v$ denote the all-one vector. The all-one matrix acts by multiplication by $v$ on $\operatorname{span}\{\mathbb{1}\}$ and vanishes on the zero-sum hyperplane, which gives the two asserted eigenvalues. The nondegeneracy hypothesis $1<k<v$ is part of the stated design setting; the eigenvalue calculation itself uses only the incidence counts encoded by $r$ and $\lambda$. [/proofplan]

[step:Compute the point Gram matrix from incidence counts]Let $\mathcal P=\{p_1,\dots,p_v\}$ denote the point set of $\mathcal D$, and let $\mathcal B=\{B_1,\dots,B_b\}$ denote the block set of $\mathcal D$, where $b=|\mathcal B|$. With respect to these enumerations, the [incidence matrix](/page/Incidence%20Matrix) $N \in \mathbb{R}^{v \times b}$ satisfies $N_{i\ell}=1$ if $p_i \in B_\ell$ and $N_{i\ell}=0$ otherwise. Let $I_v \in \mathbb{R}^{v \times v}$ denote the identity matrix, let $J \in \mathbb{R}^{v \times v}$ denote the all-one matrix, so $J_{ij}=1$ for all $1 \le i,j \le v$, and let $\mathbb{1}:=(1,\dots,1)^\top\in\mathbb{R}^v$ denote the all-one vector. For $1 \le i,j \le v$, the $(i,j)$ entry of the product $NN^\top$ is \begin{align*} (NN^\top)_{ij} = \sum_{\ell=1}^{b} N_{i\ell}N_{j\ell}. \end{align*} If $i=j$, this sum counts the blocks containing $p_i$, hence equals the [replication number](/page/Replication%20Number) $r$. If $i \ne j$, this sum counts the blocks containing both distinct points $p_i$ and $p_j$, hence equals $\lambda$ by the definition of a [$2$-$(v,k,\lambda)$ design](/page/Block%20Design). Therefore $(NN^\top)_{ij}=r$ when $i=j$, and $(NN^\top)_{ij}=\lambda$ when $i\ne j$. The matrix $(r-\lambda)I_v+\lambda J$ has the same entries, so \begin{align*} NN^\top = (r-\lambda)I_v+\lambda J. \end{align*}[/step]

[guided]We want to convert the combinatorial design axioms into a linear algebra statement about the matrix $NN^\top$. Let $\mathcal P=\{p_1,\dots,p_v\}$ denote the point set of $\mathcal D$, and let $\mathcal B=\{B_1,\dots,B_b\}$ denote the block set of $\mathcal D$, where $b=|\mathcal B|$. With respect to these enumerations, the incidence matrix $N \in \mathbb{R}^{v \times b}$ satisfies $N_{i\ell}=1$ if $p_i \in B_\ell$ and $N_{i\ell}=0$ otherwise. Let $\mathbb{1}:=(1,\dots,1)^\top\in\mathbb{R}^v$ denote the all-one vector. The entry $(NN^\top)_{ij}$ is the dot product of row $i$ and row $j$ of the incidence matrix: \begin{align*} (NN^\top)_{ij} = \sum_{\ell=1}^{b} N_{i\ell}N_{j\ell}. \end{align*} Each factor $N_{i\ell}N_{j\ell}$ equals $1$ exactly when the block $B_\ell$ contains both points $p_i$ and $p_j$, and equals $0$ otherwise. There are two cases. If $i=j$, then the sum counts blocks containing the single point $p_i$. By the definition of the [replication number](/page/Replication%20Number), this number is $r$. If $i \ne j$, then $p_i$ and $p_j$ are distinct points, and the defining property of a [$2$-$(v,k,\lambda)$ design](/page/Block%20Design) says that exactly $\lambda$ blocks contain both of them. Hence $(NN^\top)_{ij}=r$ when $i=j$, and $(NN^\top)_{ij}=\lambda$ when $i\ne j$. The matrix $(r-\lambda)I_v+\lambda J$ has diagonal entries $(r-\lambda)+\lambda=r$ and off-diagonal entries $0+\lambda=\lambda$. Since its entries agree with the entries of $NN^\top$ in every position, we obtain \begin{align*} NN^\top = (r-\lambda)I_v+\lambda J. \end{align*}[/guided]

[step:Evaluate the matrix on the all-one direction] Let $y \in \operatorname{span}\{\mathbb{1}\}$. Then there exists $c \in \mathbb R$ such that $y=c\mathbb{1}$. Since every row of $J$ has sum $v$, we have \begin{align*} J\mathbb{1}=v\mathbb{1}. \end{align*} Using the identity from the previous step, first substitute $y=c\mathbb{1}$ and $J\mathbb{1}=v\mathbb{1}$ to get \begin{align*} NN^\top y = c(r-\lambda)\mathbb{1}+c\lambda v\mathbb{1}. \end{align*} Combining the coefficients of $c\mathbb{1}$ gives \begin{align*} NN^\top y = \bigl(r+(v-1)\lambda\bigr)y. \end{align*} Thus $NN^\top$ acts on $\operatorname{span}\{\mathbb{1}\}$ by the scalar $r+(v-1)\lambda$. [/step]

[step:Evaluate the matrix on the zero-sum hyperplane] Define the zero-sum hyperplane $H := \{y \in \mathbb{R}^v : y_1+\cdots+y_v=0\}$. Let $y=(y_1,\dots,y_v)^\top \in H$. For each $1 \le i \le v$, the $i$th component of $Jy$ is \begin{align*} (Jy)_i = \sum_{j=1}^{v} y_j = 0, \end{align*} so $Jy=0$. Therefore, using $NN^\top=(r-\lambda)I_v+\lambda J$ and $Jy=0$, we obtain \begin{align*} NN^\top y = (r-\lambda)y. \end{align*} Thus $NN^\top$ acts on $H$ by the scalar $r-\lambda$. [/step]

[step:Account for all directions in $\mathbb R^v$] The subspace $\operatorname{span}\{\mathbb{1}\}$ has dimension $1$. Define the linear functional $\sigma: \mathbb{R}^v \to \mathbb{R}$ by $\sigma((y_1,\dots,y_v)^\top)=y_1+\cdots+y_v$. This functional has rank $1$, and its kernel is exactly $H$. Hence $\dim H=v-1$. Also $\operatorname{span}\{\mathbb{1}\}\cap H=\{0\}$ because if $c\mathbb{1}\in H$, then $cv=0$, so $c=0$. Therefore \begin{align*} \mathbb R^v = H \oplus \operatorname{span}\{\mathbb{1}\}. \end{align*} The preceding steps give the asserted eigenvalue of $NN^\top$ on each summand, completing the proof. [/step]